JEE Main Motion in Plane PYQ

41. (JEE Main 2013 (Offline) )

A projectile is given an initial velocity of $(\hat{i} + 2 \hat{j})$ m/s, where $\hat{i}$ is along the ground and $\hat{j}$ is along the vertical. If g = 10 m/s², the equation of its trajectory is:

A. y = x - 5x²

B. y = 2x - 5x²

C. 4y = 2x - 5x²

D. 4y = 2x - 25x²

rrect Option is (B)

$\vec{u} = \hat{i} + 2 \hat{j} = u_{x} \hat{i} + u_{y} \hat{j}$

$\Rightarrow u \cos θ = 1, u \sin θ = 2$

Also $x = u_{x} t$ and

$y = u_{y} t - \frac{1}{2} g t^{2}$

$\Rightarrow$ $y = x \tan θ - \frac{1}{2} \frac{g x^{2}}{u_{x}^{2}}$

$∴$ $y = 2 x - \frac{1}{2} g x^{2} = 2 x - 5 x^{2}$

1 42. ( AIEEE 2012 )

A boy can throw a stone up to a maximum height of 10 m. The maximum horizontal distance that the boy can throw the same stone up to will be

A. $20 \sqrt{2}$ m

B. 10 m

C. $10 \sqrt{2}$ m

D. 20 m

rrect Option is (D)

We know, $R = \frac{u^{2} \sin 2 θ}{g}$ and $H = \frac{u^{2} \sin^{2} θ}{2 g};$

$H_{max}$ is possible when $θ = 90$ $^{\circ}$

$H_{max} = \frac{u^{2}}{2 g} = 10 \Rightarrow u^{2} = 10 g \times 2$

As $R = \frac{u^{2} \sin 2 θ}{g}$

Range is maximum when projectile is thrown at an angle $45^{\circ}$ .

$\Rightarrow R_{max} = \frac{u^{2}}{g}$

$R_{max} = \frac{10 \times g \times 2}{g} = 20$ meter

2 43. (AIEEE 2011 )

A water fountain on the ground sprinkles water all around it. If the speed of water coming out of the fountain is v, the total area around the fountain that gets wet is :

A. $π \frac{v^{4}}{g^{2}}$

B. $\frac{π}{2} \frac{v^{4}}{g^{2}}$

C. $π \frac{v^{2}}{g^{2}}$

D. $π \frac{v^{2}}{g}$

rrect Option is (A)

Maximum range of water coming out of fountain,

$R_{max} = \frac{v^{2} \sin 2 θ}{g} = \frac{v^{2} \sin 90^{\circ}}{g} = \frac{v^{2}}{g}$

Total area around fountain,

$A = π R_{max}^{2} = π \frac{v^{4}}{g^{2}}$

5 44. (AIEEE 2004 )

A projectile can have the same range 'R' for two angles of projection. If T₁ and T₂ be the time of flights in the two cases, then the product of the two time of flights is directly proportional to

A. R

B. $\frac{1}{R}$

C. $\frac{1}{R^{2}}$

D. $R^{2}$

rrect Option is (A)

Range is same for angle of projection $θ,$ and $90^{\circ} - θ$

$T_{1} = \frac{2 u \sin θ}{g}, T_{2} = \frac{2 u \cos θ}{g}$

$T_{1} T_{2} =$ $\frac{4 u^{2} \sin θ \cos θ}{g^{2}}$

= $\frac{2}{g} \times (\frac{u^{2} \sin 2 θ}{g})$

= $\frac{2 R}{g}$

(as $R =$ $\frac{u^{2} \sin 2 θ}{g}$ )

Hence, $T_{1} T_{2}$ is proportional to $R .$

7 45. (AIEEE 2004 )

A ball is thrown from a point with a speed ν₀ at an angle of projection θ. From the same point and at the same instant person starts running with a constant speed $\frac{v_{0}}{2}$ to catch the ball. Will the person be able to catch the ball? If yes, what should be the angle of projection θ?

A. No

B. Yes, $30^{\circ}$

C. Yes, $60^{\circ}$

D. Yes, $45^{\circ}$

rrect Option is (C)

Yes, the person can catch the ball when horizontal velocity is equal to the horizontal component of ball's velocity, the motion of ball will be only in vertical direction with respect to person for that,

$\frac{v_{0}}{2} = v_{0} \cos θ$

or $\cos θ = \frac{1}{2}$

$\Rightarrow \cos θ = \cos 60^{\circ}$

$\Rightarrow θ = 60^{\circ}$